Explanation 1) Concept: We use the condition of constructive interference to find the expression for thickness of the film. We can also find the thickness from the given figure. By equating these two relations, we can find the radius of the interference maxima. Using the expression for radius, we can find the number of bright fringes m for air and water. By replacing m by ( m + 20 ) and subtracting the two relations, we can get the radius of curvature of the lower surface of thelens. 2) Formula: 2 L = m + 1 2 λ n 3) Given: i) A Newton’s rings apparatus is to be used to determine the radius of curvature of a lens. ii) Refer to Fig.35.46 and Problem 75 of the textbook. iii) Radius of m th bright fringe is r = 0.162 c m iv) Radius of ( m + 20 ) th bright fringe is r ' = 0.368 c m v) Wavelength of light in air λ = 546 n m 4) Calculations: Consider the interferance pattern formed by the waves reflected from the upper and lower surface of the air wedge. The wave reflected from the upper surface has no phase shift while the wave reflected from the lower surface has the phase shift of π radians. At thickness d of the wedge, the condition for maximum in intensity is 2 d = m + 1 2 λ n where, n is the refractive index of air, i.e., n = 1 2 d = m + 1 2 λ And λ is the wavelength in the air and m is an integer. So we get d = 2 m + 1 λ 4 … … … … … … … … … … … … … … … … … … … … ( 1 ) From figure 35 - 46 ( a ) , using geometry, d = R - R 2 - r 2 … … … … … … … … … … … … … … … … … … … … ( 2 ) where R is the radius of curvature of the lens and r is the radius of Newton’s ring. From equations (1) and (2), we get 2 m + 1 λ 4 = R - R 2 - r 2 R 2 - r 2 = R - 2 m + 1 λ 4 Squaring both sides and solving for radius r , R 2 - r 2 = R 2 + 2 m + 1 2 λ 4 2 - 2 m + 1 R λ 2 r 2 = 2 m + 1 R λ 2 - 2 m + 1 2 λ 4 2 r = 2 m + 1 R λ 2 - 2 m + 1 2 λ 2 16 If the radius of curvature of the ring is much greater than the wavelength of light, i

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Chapter 35, Problem 77P

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the radius of curvature of the lower surface of the lens.

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Chapter 35, Problem 76P

Chapter 35, Problem 78P

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Chapter 35 Solutions

Fundamentals Of Physics - Volume 1 Only

Ch. 35 - Does the spacing between fringes in a two-slit...Ch. 35 - a If you move from one bright fringe in a two-slit...Ch. 35 - Figure 35-22 shows two light rays that are...Ch. 35 - In Fig. 35-23, three pulses of lighta, b, and cof...Ch. 35 - Is there an interference maximum, a minimum, an...Ch. 35 - Figure 35-24a gives intensity I verus position x...Ch. 35 - Figure 35-25 shows two sources S1 and S2 that emit...Ch. 35 - Figure 35-26 shows two rays of light, of...Ch. 35 - Light travels along the length of a 1500-nm-long...Ch. 35 - Figure 35-27a shows the cross section of a...

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the radius of curvature of the lower surface of the lens.

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To find: the radius of curvature of the lower surface of the lens.

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Chapter 35 Solutions

To find:

the radius of curvature of the lower surface of the lens.